Answer
$-0.7831$
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$|R_n| \leq \dfrac{1}{(n+1)^{n+1}} \lt 10^{-3}$
and $ (n+1)^{n+1} \gt 10^3$
or, $n +1=5$
or, $n =4$
This implies that the value for $n$ is $n=4$ and the remainder is less than $10^{-3}$.
Next, the value of series is: $\sum_{k=1}^{\infty} \dfrac{(-1)^k}{k^k}=-1+\dfrac{1}{2^2}-\dfrac{1}{3^3}+\dfrac{1}{4^4}=-0.7831$
